Question
An 800.0 kg roller coaster car is at rest at the top of a 95 m hill. It rolls down the first drop to a height of 31 m. When it travels to the top of the second hill, it is moving at 28 m/s. It then rolls down the second hill until it is at ground level.
What is the kinetic and potential energy at the top and bottom of each hill?
What is the kinetic and potential energy at the top and bottom of each hill?
Answers
They presumably want to to assume total mechanical energy (kinetic + potential) is constant, even though that is not the case.
At the top of the 95 m hill, all of the energy is potential, and equals
M g H = 744,800 J. Use the height or velocity at the other location, and the total energy (748,800 J) , to determine kinetic and potential energies
At the top of the 95 m hill, all of the energy is potential, and equals
M g H = 744,800 J. Use the height or velocity at the other location, and the total energy (748,800 J) , to determine kinetic and potential energies
1.
PE1 = mgh1 = 800•9.8•95 =744800 J.
KE1=0.
Total E1 = PE1+ KE1=744800 J.
2.
PE2 = 800•9.8•31=243040
PE1= PE2+KE2
KE2 = PE1- PE2 =
=744800 - 243040=501760 J.
Total E2 = PE2+ KE2=744800 J.
3.
KE3 =mv²/2= 800•(28)²/2 =313600 J.
PE3 =744800-313600 = 431200 J.
Total E3 = PE3+ KE3=744800 J.
4.
PE4=0.
KE4 = 744800 J.
Total E4 = PE4+ KE4=744800 J.
PE1 = mgh1 = 800•9.8•95 =744800 J.
KE1=0.
Total E1 = PE1+ KE1=744800 J.
2.
PE2 = 800•9.8•31=243040
PE1= PE2+KE2
KE2 = PE1- PE2 =
=744800 - 243040=501760 J.
Total E2 = PE2+ KE2=744800 J.
3.
KE3 =mv²/2= 800•(28)²/2 =313600 J.
PE3 =744800-313600 = 431200 J.
Total E3 = PE3+ KE3=744800 J.
4.
PE4=0.
KE4 = 744800 J.
Total E4 = PE4+ KE4=744800 J.
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